Updated spacing between @ latex expressions

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revision #10366 parent #9214 GitHub c702ebe ← older newer →
en/4.2.26.md +4 −4
@@ -80,24 +80,24 @@
<p>
Since the ball does not float, the vertical Archimedean force is completely compensated by the vertical component of the thread tension force
−$$T\cos\alpha=\frac{4\pi r^3}{3}\rho g$$
+$$T\cos\alpha =\frac{4\pi r^3}{3}\rho g$$
From where we could find the thread tension force
$$T=\frac{4\pi r^3}{3\cos\alpha}\rho g\quad (1)$$
Alternatively, we could write an equilibrium condition for the <b>Horizontal Archimedean force</b>, which is compensated by the horizontal force of the thread tension
$$T\sin\alpha = \rho V\omega^2x\quad (2)$$
It can be seen from the figure that the ball's center of mass is at a distance $l+r$ from the point of attachment of the thread:
$$x=R-(l+r)\sin\alpha$$
After substituting into $(2)$ we obtain
−$$\rho \frac{4\pi r^3}{3}\omega^2\left(R-(l+r)\sin\alpha\right) = T\sin\alpha \quad (3)$$
+$$\rho\frac{4\pi r^3}{3}\omega^2\left(R-(l+r)\sin\alpha\right) = T\sin\alpha\quad (3)$$
Substituting $(1)$ into $(3)$
−<div class="scroll-wrapper">$$\rho \frac{4\pi r^3}{3}\omega^2\left(R-(l+r)\sin\alpha\right) = \frac{4\pi r^3}{3\cos\alpha}\rho g\sin\alpha$$</div>
+<div class="scroll-wrapper">$$\rho\frac{4\pi r^3}{3}\omega^2\left(R-(l+r)\sin\alpha\right) = \frac{4\pi r^3}{3\cos\alpha}\rho g\sin\alpha$$</div>
From where we find the angular velocity
$$\boxed{\omega = \sqrt{\frac{g\tan\alpha}{R-(l+r)\sin\alpha}}}$$
</p>
<h4>Answer</h4>
<p>
− $$\omega=\sqrt{(g\operatorname{tg}\alpha)/[R-(l+r)\operatorname{sin}\alpha]}.$$
+ $$\omega =\sqrt{(g\operatorname{tg}\alpha )/[R-(l+r)\operatorname{sin}\alpha ]}.$$
</p>
<p style="text-align: right; font-style: italic; font-size: 14;">
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